# How to show that the initial value problem has a unique solution in the given interval?

Use Picard’s theorem to show that the initial value problem $$(1+e^x)\frac{dy}{dx} = \sin(x + y^3)$$, $$y(1) = 3$$, has a unique solution on the interval $$x ≥ 1$$.

By Picard's Existence and Uniqueness Theorem; If $$f$$ is continuous on a domain $$D$$ and $$f$$ satisfies Lipschitz condition on $$D$$. If $$R=\{|x−x_0|≤a;\,|y−y_0|≤b\}$$ lies in $$D$$ and $$M={\rm maximum}|f(x,y)|$$, $$\alpha=\min\{a,b/M\}$$. Then the IVP has a unique solution on the interval $$|x−x_0|≤\alpha$$.

I'm not sure how to find the value of M and alpha and then go on to find the interval.

I've figured out that M=1/(1+e^(1-a)) and alpha= min(a,b(1+e^(1-a)) however I cant figure out a and b.

For existence theorem to be used f(x,y) needs to be continuous in the interval $$\begin{equation} R=\left\{(x, y):\left|x-x_{0}\right| \leq a,\left|y-y_{0}\right| \leq b\right\}, \quad(a, b>0) \end{equation}$$ To find the interval actually you should randomly pick the area by yourself because you are analysing the equation and you pick those values of a and b according to the function so, you should pick some interval and test it with the existence and uniqueness theorems but they are not fully telling you the interval of validity so you should find the interval first according to some techniques.